How do you find the volume of the solid #y=sqrt(9-x^2)# revolved about the x-axis?
Volume
graph{(y-sqrt(9-x^2))=0 [-6, 6, -2, 4]}
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To find the volume of the solid generated by revolving the curve ( y = \sqrt{9 - x^2} ) about the x-axis, you can use the method of cylindrical shells or the disk/washer method.
Using the disk/washer method, you integrate from ( x = -3 ) to ( x = 3 ) (since ( y = \sqrt{9 - x^2} ) is defined for ( -3 \leq x \leq 3 )) and evaluate the integral:
[ V = \pi \int_{-3}^{3} (\sqrt{9 - x^2})^2 , dx ]
This integral represents the volume of infinitely thin disks (or washers) at each value of ( x ), summed together over the interval ( [-3, 3] ).
After integrating, you will find the volume of the solid generated by revolving ( y = \sqrt{9 - x^2} ) about the x-axis.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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